This module extends the ideas of the Discrete Fourier Transform (DFT) into two-dimensions, which is necessary for any image processing.
2d dft
To perform image restoration (and many other useful image
processing algorithms) in a computer, we need a FourierTransform (FT) that is discrete and two-dimensional.
where the above equation (
)
has finite support for an
$N$ x
$N$ image.
Inverse 2d dft
As with our regular fourier transforms, the 2D DFT also has
an inverse transform that allows us to reconstruct an imageas a weighted combination of complex sinusoidal basis
functions.
Below we go through the steps of convolving two
two-dimensional arrays. You can think of
$f$ as representing an image and
$h$ represents a PSF, where
$h(m, n)=0$ for
$m\land n> 1$ and
$m\land n< 0$ .
$$h=\begin{pmatrix}h(0, 0) & h(0, 1)\\ h(1, 0) & h(1, 1)\\ \end{pmatrix}$$$$f=\begin{pmatrix}f(0, 0) & & f(0, N-1)\\ & & \\ f(N-1, 0) & & f(N-1, N-1)\\ \end{pmatrix}$$ Step 1 (Flip
$h$ ):
We use the standard 2D convolution equation (
) to find the first element of
our convolved image. In order to better understand what ishappening, we can think of this visually. The basic idea is
to take
$h(-m, -n)$ and place it "on top" of
$f(k, l)$ , so that just the bottom-right element,
$h(0, 0)$ of
$h(-m, -n)$ overlaps with the top-left element,
$f(0, 0)$ , of
$f(k, l)$ . Then, to get the next element of our convolved
image, we slide the flipped matrix,
$h(-m, -n)$ , over one element to the right and get the
following result:
$$g(0, 1)=h(0, 0)f(0, 1)+h(0, 1)f(0, 0)$$ We continue in this fashion to find all of the elements ofour convolved image,
$g(m, n)$ . Using the above method we define the general
formula to find a particular element of
$g(m, n)$ as:
Using this equation, we can calculate the value for each
position on our final image,
$\stackrel{~}{g}(m, n)$ . For example, due to the periodic extension of
the images, when circular convolution is applied we willobserve a
wrap-around effect.
where in the above equation,
$\sum_{n=0}^{N-1} f(m, n)e^{-i\frac{2\pi ln}{N}}$ is simply a 1D DFT over
$n$ .
This means that we will take the 1D FFT of each row; if wehave
$N$ rows, then it will
require
$N\lg N$ operations per row. We can rewrite this as
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